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$$\int_{-\infty}^{+\infty} F(x-b) f(x-a) dx$$

$$f(x) = \exp(-x-e^{-x}), \qquad x \in (-\infty, +\infty)$$

$$F(x) = \int_{-\infty}^x f(t) dt$$

I calculated already the integral of $F(x)$, which is $\exp(-e^{-x})$, but I am stuck on the other one, I have no idea how to calculate the integral of $F(x-a)f(x-b)$.

Rócherz
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user10471408
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1 Answers1

1

$$\int_{-\infty}^{\infty}F(a-b)f(x-a)dx=\int_{-\infty}^{\infty}e^{-e^{-(x-b)}} e^{-(x-a)} e^{-e^{-(x-a)}} dx=$$

$\Rightarrow e^{a} \int_{-\infty}^{\infty} e^{-e^{-x}({e^{b}+e^{a}})} e^{-x} dx$

Now Let, $-e^{-x}=t, \rightarrow e^{-x} dx =dt$

$\Rightarrow e^{a} \int_{-\infty}^{0} e^{t{(e^{a}+e^{b}})} dt~~$

$\Rightarrow \frac{e^a}{e^{a}+e^{b}}.$

Schrodinger
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